1. Find the direction cosines of the vector 2i + 2j – k. 2. ​ Find the distance of the point (2, 3, 4) from the plane

Question

1. Find the direction cosines of the vector 2i + 2j – k.
2. Find the distance of the point (2, 3, 4) from the plane \(\bar r\).(3i – 6j + 2 k) = -11.
3. Find the shortest distance between the lines \(\bar r\) = (2i – j – k)+ λ(3i – 5 j + 2k) an \(\bar r\) = (i+ 2 j + k)+ µ(i – j + k)

Answer 1

1. Direction ratios of the vector 2i + 2j – k is 2, 2, -1

Direction cosines of the vector 2i + 2j – k is

2. The equation of the plane in the Cartesian form is 3x – 6y + 2z + 11 = 0 . Then distance from the point (2, 3, 4) is

3. The given lines are \(\bar r\) = (2i – j – k) + λ(3i – 5j + 2k)